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Level 5Grade 5skill: multi_digit_mult· 8 min read

Long Multiplication (Multi-Digit)

In short: Multiply the top number by one bottom digit at a time, giving each result its own row. Shift every new row one place left (drop a 0) because that digit is really worth tens or hundreds, then add all the rows. Master carries, the left shift, and adding every partial product and you can multiply any two whole numbers.

How to multiply two multi-digit numbers

Long multiplication looks like a lot of steps, but it is really one idea repeated: you multiply the top number by one digit at a time, then add up what you get. The only thing you have to respect is place value — which column each answer belongs in.

Write the number with more digits on top and line the columns up neatly. Work the bottom digits from right to left, one at a time. When a single fact spills past 9, carry the extra tens into the next column and add them there.

  1. Write the larger number on top and line up the digits by place value.
  2. Multiply the top number by the ones digit of the bottom number, carrying whenever a fact goes past 9. This is your first row.
  3. Put a 0 in the ones column of the next row, then multiply the top number by the tens digit. That 0 is what shifts the row one place to the left.
  4. Add one more row (and one more 0) for every extra digit in the bottom number.
  5. Add all the partial products together. That total is your answer.

Why the shifting zero is the whole trick Why it works

Split the multiplier
Break the bottom number into its place-value parts and multiply by each part separately.

253 × 46 = 253 × 40 + 253 × 6. Those two pieces are the entire problem — there is no third row to invent and none to skip.

The row that shifts
Each new row starts one column further left, because that digit is worth ten times more than the one before it.

The second row of 253 × 46 is 253 × 40 = 10120, so it ends in 0. Sliding the row left is that 0.

Carry, don't lose it
When one single-digit fact goes past 9, the extra tens ride into the next column and get added there.

In 253 × 6, the fact 5 × 6 = 30 carries a 3, so the next column 2 × 6 = 12 becomes 15 — giving 1518, not 1508.

Here is the one thing worth remembering: the blank space (or the 0) under the second row is not decoration and it is not optional. So use it as a check — if your final total ever looks far too small for the numbers you started with, the shift is the first thing to inspect, because a row that forgot to slide left is quietly answering an easier question than the one you were asked.

Watch it work, row by row

47 × 23 = 1081
  1. Ones row: 47 × 3. 7×3 = 21, write 1 carry 2; 4×3 = 12, +2 = 14. Row = 141.
  2. Tens row: put a 0, then 47 × 2 = 94, so the row is 940.
  3. Add: 141 + 940 = 1081.
68 × 54 = 3672
  1. Ones row: 68 × 4. 8×4 = 32, write 2 carry 3; 6×4 = 24, +3 = 27. Row = 272.
  2. Tens row: put a 0, then 68 × 5 = 340, so the row is 3400.
  3. Add: 272 + 3400 = 3672.
306 × 27 (watch the 0 in the middle) = 8262
  1. Ones row: 306 × 7. 7×6 = 42, write 2 carry 4; 7×0 = 0, +4 = 4; 7×3 = 21. Row = 2142.
  2. Tens row: put a 0, then 306 × 2 = 612, so the row is 6120.
  3. Add: 2142 + 6120 = 8262. The 0 in 306 still gets multiplied — it just contributes 0 plus any carry.
253 × 46 = 11638
  1. Ones row: 253 × 6. 3×6 = 18, write 8 carry 1; 5×6 = 30, +1 = 31, write 1 carry 3; 2×6 = 12, +3 = 15. Row = 1518.
  2. Tens row: put a 0, then 253 × 4 = 1012, so the row is 10120.
  3. Add: 1518 + 10120 = 11638.

Four ways this goes wrong

Reaching for the wrong times-table fact

✗ 11698

The 5 and 6 rows of the six-times table sit right next to each other — 5 × 6 = 30 and 6 × 6 = 36 — so it is easy to grab the neighbour and use 36 where you needed 30. That turns the first row of 253 × 6 into 1578 instead of 1518, and 1578 + 10120 = 11698.

Fix: If a fact ever feels shaky, rebuild it from one you trust: 5 × 6 is one group of 6 less than 6 × 6 = 36, so it is 30. One wrong fact poisons a whole row.

table_neighbour

Forgetting to shift the second row

✗ 2530

The tens row of 253 × 46 is 253 × 40 = 10120, not 253 × 4 = 1012. Skip the shifting 0 and you write 1012 in the ones column, so you add 1518 + 1012 = 2530 — an answer far too small, because you multiplied by 4 instead of 40.

Fix: Always drop the 0 (or leave a clear blank) before the second row starts. If your final answer looks too small for the numbers, a lost shift is the first thing to check.

place_value_slip

Stopping after the first row

✗ 1518

It is tempting to finish 253 × 6 = 1518, feel done, and hand that in — but you have only multiplied by the 6 and completely forgotten the 253 × 40 row. You added no second partial product at all.

Fix: Count your rows before you add: a two-digit multiplier must give you two rows, a three-digit one gives three. If you have fewer rows than the bottom number has digits, you stopped early.

partial_computation

Practice

Work each one on paper, row by row, and keep your columns lined up. Watch your carries and never forget the shifting 0 on the second row.

Before and after this skill

First, make sure of

Frequently asked questions

Which number should go on top?

Mathematically it does not matter — 47 × 23 and 23 × 47 give the same answer. But put the number with more digits on top: the bottom number decides how many rows you write, so a shorter bottom number means less work and fewer chances to slip.

Do I write the 0 on the second row, or just leave a blank?

Either is correct, because a blank column and a 0 mean the same thing here. Writing the 0 is safer while you are learning — it physically forces the row into the right column so you cannot forget the shift.

What do I do with a 0 inside a number, like the 0 in 306?

You still multiply it. Any digit times 0 is 0, but you must add in any carry from the column before. In 306 × 7, the middle step is 7 × 0 = 0, plus the carried 4, which gives 4 — so the row is 2142, not 242.

How can I check I haven't lost a row?

Count the digits in the bottom number: that is exactly how many rows you should add. A two-digit multiplier gives two rows, a three-digit multiplier gives three. Fewer rows than digits means you stopped a step early.